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This commit is contained in:
CyC2018
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@ -349,7 +349,8 @@ public int findContentChildren(int[] g, int[] s) {
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Arrays.sort(s);
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int gIndex = 0, sIndex = 0;
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while (gIndex < g.length && sIndex < s.length) {
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if (g[gIndex] <= s[sIndex]) gIndex++;
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if (g[gIndex] <= s[sIndex])
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gIndex++;
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sIndex++;
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}
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return gIndex;
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@ -367,11 +368,10 @@ public int findContentChildren(int[] g, int[] s) {
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```java
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public int maxProfit(int[] prices) {
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int profit = 0;
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for (int i = 1; i < prices.length; i++) {
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if (prices[i] > prices[i - 1]) {
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for (int i = 1; i < prices.length; i++)
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if (prices[i] > prices[i - 1])
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profit += (prices[i] - prices[i - 1]);
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}
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}
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return profit;
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}
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```
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@ -389,11 +389,13 @@ Output: True
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```java
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public boolean canPlaceFlowers(int[] flowerbed, int n) {
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int len = flowerbed.length;
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int cnt = 0;
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for (int i = 0; i < flowerbed.length; i++) {
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if (flowerbed[i] == 1) continue;
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for (int i = 0; i < len; i++) {
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if (flowerbed[i] == 1)
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continue;
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int pre = i == 0 ? 0 : flowerbed[i - 1];
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int next = i == flowerbed.length - 1 ? 0 : flowerbed[i + 1];
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int next = i == len - 1 ? 0 : flowerbed[i + 1];
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if (pre == 0 && next == 0) {
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cnt++;
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flowerbed[i] = 1;
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@ -415,17 +417,19 @@ Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
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题目描述：判断一个数组能不能只修改一个数就成为非递减数组。
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在出现 nums[i] < nums[i - 1] 时，需要考虑的是应该修改数组的哪个数，使得本次修改能使 i 之前的数组成为非递减数组，并且 **不影响后续的操作** 。优先考虑令 nums[i - 1] = nums[i]，因为如果修改 nums[i] = nums[i - 1] 的话，那么 nums[i] 这个数会变大，就有可能比 nums[i + 1] 大，从而影响了后续操作。还有一个比较特别的情况就是 nums[i] < nums[i - 2]，只修改 nums[i - 1] = nums[i] 不能令数组成为非递减，只能通过修改 nums[i] = nums[i - 1] 才行。
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在出现 nums[i] < nums[i - 1] 时，需要考虑的是应该修改数组的哪个数，使得本次修改能使 i 之前的数组成为非递减数组，并且 **不影响后续的操作** 。优先考虑令 nums[i - 1] = nums[i]，因为如果修改 nums[i] = nums[i - 1] 的话，那么 nums[i] 这个数会变大，就有可能比 nums[i + 1] 大，从而影响了后续操作。还有一个比较特别的情况就是 nums[i] < nums[i - 2]，只修改 nums[i - 1] = nums[i] 不能使数组成为非递减数组，只能修改 nums[i] = nums[i - 1]。
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```java
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public boolean checkPossibility(int[] nums) {
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int cnt = 0;
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for (int i = 1; i < nums.length; i++) {
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if (nums[i] < nums[i - 1]) {
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cnt++;
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if (i - 2 >= 0 && nums[i - 2] > nums[i]) nums[i] = nums[i - 1];
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else nums[i - 1] = nums[i];
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}
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for (int i = 1; i < nums.length && cnt < 2; i++) {
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if (nums[i] >= nums[i - 1])
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continue;
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cnt++;
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if (i - 2 >= 0 && nums[i - 2] > nums[i])
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nums[i] = nums[i - 1];
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else
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nums[i - 1] = nums[i];
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}
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return cnt <= 1;
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}
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@ -442,15 +446,61 @@ Return true.
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```java
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public boolean isSubsequence(String s, String t) {
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int pos = -1;
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int index = -1;
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for (char c : s.toCharArray()) {
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pos = t.indexOf(c, pos + 1);
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if (pos == -1) return false;
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index = t.indexOf(c, index + 1);
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if (index == -1)
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return false;
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}
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return true;
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}
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```
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**不重叠的区间个数**
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[435. Non-overlapping Intervals (Medium)](https://leetcode.com/problems/non-overlapping-intervals/description/)
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```html
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Input: [ [1,2], [1,2], [1,2] ]
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Output: 2
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Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
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```
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```html
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Input: [ [1,2], [2,3] ]
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Output: 0
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Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
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```
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题目描述：计算让一组区间不重叠所需要移除的区间个数。
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直接计算最多能组成的不重叠区间个数即可。
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在每次选择中，区间的结尾最为重要，选择的区间结尾越小，留给后面的区间的空间越大，那么后面能够选择的区间个数也就越大。
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按区间的结尾进行排序，每次选择结尾最小，并且和前一个区间不重叠的区间。
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```java
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public int eraseOverlapIntervals(Interval[] intervals) {
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if (intervals.length == 0)
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return 0;
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Arrays.sort(intervals, Comparator.comparingInt(o -> o.end));
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int cnt = 1;
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int end = intervals[0].end;
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for (int i = 1; i < intervals.length; i++) {
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if (intervals[i].start < end)
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continue;
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end = intervals[i].end;
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cnt++;
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}
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return intervals.length - cnt;
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}
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```
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**投飞镖刺破气球**
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[452. Minimum Number of Arrows to Burst Balloons (Medium)](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/description/)
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@ -465,31 +515,27 @@ Output:
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题目描述：气球在一个水平数轴上摆放，可以重叠，飞镖垂直投向坐标轴，使得路径上的气球都会刺破。求解最小的投飞镖次数使所有气球都被刺破。
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对气球按末尾位置进行排序，得到：
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```html
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[[1,6], [2,8], [7,12], [10,16]]
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```
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如果让飞镖投向 6 这个位置，那么 [1,6] 和 [2,8] 这两个气球都会被刺破，这种方式下刺破这两个气球的投飞镖次数最少，并且后面两个气球依然可以使用这种方式来刺破。
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也是计算不重叠的区间个数，不过和 Non-overlapping Intervals 的区别在于，[1, 2] 和 [2, 3] 在本题中算是重叠区间。
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```java
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public int findMinArrowShots(int[][] points) {
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if (points.length == 0) return 0;
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Arrays.sort(points, (a, b) -> (a[1] - b[1]));
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int curPos = points[0][1];
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int shots = 1;
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if (points.length == 0)
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return 0;
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Arrays.sort(points, Comparator.comparingInt(o -> o[1]));
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int cnt = 1, end = points[0][1];
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for (int i = 1; i < points.length; i++) {
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if (points[i][0] <= curPos) {
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if (points[i][0] <= end) // [1,2] 和 [2,3] 算重叠
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continue;
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}
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curPos = points[i][1];
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shots++;
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cnt++;
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end = points[i][1];
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}
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return shots;
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return cnt;
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}
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```
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**分隔字符串使同种字符出现在一起**
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[763. Partition Labels (Medium)](https://leetcode.com/problems/partition-labels/description/)
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@ -504,25 +550,25 @@ A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits
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```
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```java
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public List<Integer> partitionLabels(String S) {
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List<Integer> partitions = new ArrayList<>();
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int[] lastIndexs = new int[26];
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for (int i = 0; i < S.length(); i++) {
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lastIndexs[S.charAt(i) - 'a'] = i;
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}
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int firstIndex = 0;
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while (firstIndex < S.length()) {
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int lastIndex = firstIndex;
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for (int i = firstIndex; i < S.length() && i <= lastIndex; i++) {
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int index = lastIndexs[S.charAt(i) - 'a'];
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if (index == i) continue;
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if (index > lastIndex) lastIndex = index;
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}
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partitions.add(lastIndex - firstIndex + 1);
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firstIndex = lastIndex + 1;
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}
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return partitions;
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}
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public List<Integer> partitionLabels(String S) {
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int[] lastIndexs = new int[26];
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for (int i = 0; i < S.length(); i++)
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lastIndexs[S.charAt(i) - 'a'] = i;
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List<Integer> ret = new ArrayList<>();
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int firstIndex = 0;
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while (firstIndex < S.length()) {
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int lastIndex = firstIndex;
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for (int i = firstIndex; i < S.length() && i <= lastIndex; i++) {
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int index = lastIndexs[S.charAt(i) - 'a'];
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if (index > lastIndex)
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lastIndex = index;
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}
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ret.add(lastIndex - firstIndex + 1);
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firstIndex = lastIndex + 1;
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}
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return ret;
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}
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```
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**根据身高和序号重组队列**
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@ -545,25 +591,17 @@ Output:
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```java
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public int[][] reconstructQueue(int[][] people) {
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if (people == null || people.length == 0 || people[0].length == 0) return new int[0][0];
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Arrays.sort(people, (a, b) -> {
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if (a[0] == b[0]) return a[1] - b[1];
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return b[0] - a[0];
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});
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int N = people.length;
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List<int[]> tmp = new ArrayList<>();
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for (int i = 0; i < N; i++) {
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int index = people[i][1];
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int[] p = new int[]{people[i][0], people[i][1]};
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tmp.add(index, p);
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}
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if (people == null || people.length == 0 || people[0].length == 0)
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return new int[0][0];
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int[][] ret = new int[N][2];
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for (int i = 0; i < N; i++) {
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ret[i][0] = tmp.get(i)[0];
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ret[i][1] = tmp.get(i)[1];
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}
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return ret;
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Arrays.sort(people, (a, b) -> (a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]));
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List<int[]> queue = new ArrayList<>();
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for (int[] p : people)
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queue.add(p[1], p);
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return queue.toArray(new int[queue.size()][]);
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}
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```
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@ -466,7 +466,7 @@ Redis 没有关系型数据库中的表这一概念来将同类型的数据存
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# 参考资料
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- Carlson J L. Redis in Action[J]. Media.johnwiley.com.au, 2013.
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- 黄健宏. Redis 设计与实现 [M]. 机械工业出版社, 2014.
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- [黄健宏. Redis 设计与实现 [M]. 机械工业出版社, 2014.](http://redisbook.com/index.html)
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- [REDIS IN ACTION](https://redislabs.com/ebook/foreword/)
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- [论述 Redis 和 Memcached 的差异](http://www.cnblogs.com/loveincode/p/7411911.html)
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- [Redis 3.0 中文版- 分片](http://wiki.jikexueyuan.com/project/redis-guide)
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@ -115,7 +115,8 @@ position-4 : (0,1,2,3,2,5) // nums[i] == nums[nums[i]], exit
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```java
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public boolean duplicate(int[] nums, int length, int[] duplication) {
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if (nums == null || length <= 0) return false;
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if (nums == null || length <= 0)
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return false;
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for (int i = 0; i < length; i++) {
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while (nums[i] != i) {
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if (nums[i] == nums[nums[i]]) {
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@ -1,5 +1,5 @@
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<!-- GFM-TOC -->
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* [一、 概述](#一-概述)
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* [一、概述](#一概述)
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* [操作系统基本特征](#操作系统基本特征)
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* [操作系统基本功能](#操作系统基本功能)
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* [系统调用](#系统调用)
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@ -31,13 +31,13 @@
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<!-- GFM-TOC -->
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# 一、 概述
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# 一、概述
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## 操作系统基本特征
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### 1. 并发
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并发性是指宏观上在一段时间内能同时运行多个程序，而并行性则指同一时刻能运行多个指令。
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并发是指宏观上在一段时间内能同时运行多个程序，而并行则指同一时刻能运行多个指令。
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并行需要硬件支持，如多流水线或者多处理器。
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@ -45,7 +45,7 @@
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### 2. 共享
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共享是指系统中的资源可以供多个并发进程共同使用。
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共享是指系统中的资源可以被多个并发进程共同使用。
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有两种共享方式：互斥共享和同时共享。
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@ -69,15 +69,17 @@
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### 2. 内存管理
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内存分配、地址映射、内存保护与共享和内存扩充等功能。
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内存分配、地址映射、内存保护与共享、内存扩充等。
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### 3. 文件管理
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文件存储空间的管理、目录管理及文件读写管理和保护等。
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文件存储空间的管理、目录管理、文件读写管理和保护等。
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### 4. 设备管理
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完成用户的 I/O 请求，方便用户使用各种设备，并提高设备的利用率，主要包括缓冲管理、设备分配、设备处理和虛拟设备等功能。
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完成用户的 I/O 请求，方便用户使用各种设备，并提高设备的利用率。
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主要包括缓冲管理、设备分配、设备处理、虛拟设备等。
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## 系统调用
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Reference in New Issue
Block a user