标签：desc interval nat 更新 HERE val lan style i++
问题：
给定数组，切分为left和right，使得left的所有元素<=right的所有元素，返回left的长度
Example 1: Input: [5,0,3,8,6] Output: 3 Explanation: left = [5,0,3], right = [8,6] Example 2: Input: [1,1,1,0,6,12] Output: 4 Explanation: left = [1,1,1,0], right = [6,12] Note: 2 <= A.length <= 30000 0 <= A[i] <= 10^6 It is guaranteed there is at least one way to partition A as described.
解法：
遍历数组，记录left的最大值maxleft
若遍历到的A[i]<maxleft，那么left数组要扩张到A[i]，
这时候maxleft也要更新到A[0]~A[i]的最大值maxnow。
代码参考：
1 class Solution { 2 public: 3 int partitionDisjoint(vector<int>& A) { 4 int maxleft=A[0], maxnow=A[0], res=0; 5 for(int i=1; i<A.size(); i++){ 6 if(A[i]<maxleft) { 7 res=i; 8 maxleft=maxnow; 9 }else{ 10 maxnow=max(maxnow, A[i]); 11 } 12 } 13 return res+1; 14 } 15 };
915. Partition Array into Disjoint Intervals
标签：desc interval nat 更新 HERE val lan style i++
原文地址：https://www.cnblogs.com/habibah-chang/p/12941879.html
