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CyC2018 2018-06-12 11:12:01 +08:00
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notes/Leetcode 题解.md
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@ -2587,26 +2587,21 @@ Explanation: The longest chain is [1,2] -> [3,4]
```java ```java
public int findLongestChain(int[][] pairs) { public int findLongestChain(int[][] pairs) {
if(pairs == null || pairs.length == 0) { if (pairs == null || pairs.length == 0) {
return 0; return 0;
} }
Arrays.sort(pairs, (a, b) -> (a[0] - b[0])); Arrays.sort(pairs, (a, b) -> (a[0] - b[0]));
int n = pairs.length; int n = pairs.length;
int[] dp = new int[n]; int[] dp = new int[n];
Arrays.fill(dp, 1); Arrays.fill(dp, 1);
for(int i = 0; i < n; i++) { for (int i = 1; i < n; i++) {
for(int j = 0; j < i; j++) { for (int j = 0; j < i; j++) {
if(pairs[i][0] > pairs[j][1]){ if (pairs[j][1] < pairs[i][0]) {
dp[i] = Math.max(dp[i], dp[j] + 1); dp[i] = Math.max(dp[i], dp[j] + 1);
} }
} }
} }
return Arrays.stream(dp).max().orElse(0);
int ret = 0;
for(int num : dp) {
ret = Math.max(ret, num);
}
return ret;
} }
``` ```
@ -2627,18 +2622,20 @@ Input: [1,2,3,4,5,6,7,8,9]
Output: 2 Output: 2
``` ```
要求：使用 O(n) 时间复杂度求解。 要求：使用 O(N) 时间复杂度求解。
使用两个状态 up 和 down。
```java ```java
public int wiggleMaxLength(int[] nums) { public int wiggleMaxLength(int[] nums) {
int len = nums.length; if (nums == null || nums.length == 0) {
if (len == 0) return 0; return 0;
}
int up = 1, down = 1; int up = 1, down = 1;
for (int i = 1; i < len; i++) { for (int i = 1; i < nums.length; i++) {
if (nums[i] > nums[i - 1]) up = down + 1; if (nums[i] > nums[i - 1]) {
else if (nums[i] < nums[i - 1]) down = up + 1; up = down + 1;
} else if (nums[i] < nums[i - 1]) {
down = up + 1;
}
} }
return Math.max(up, down); return Math.max(up, down);
} }
@ -2671,8 +2668,11 @@ public int lengthOfLCS(int[] nums1, int[] nums2) {
int[][] dp = new int[n1 + 1][n2 + 1]; int[][] dp = new int[n1 + 1][n2 + 1];
for (int i = 1; i <= n1; i++) { for (int i = 1; i <= n1; i++) {
for (int j = 1; j <= n2; j++) { for (int j = 1; j <= n2; j++) {
if (nums1[i - 1] == nums2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1; if (nums1[i - 1] == nums2[j - 1]) {
else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]); dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
} }
} }
return dp[n1][n2]; return dp[n1][n2];