*LeetCode-Trapping Rain Water
来源:互联网 发布:二叉树的深度 java 编辑:程序博客网 时间:2024/06/10 03:06
感觉很难想到 两个指针从两边 两边也都keep一个max 每次只挪动一个 根据当前那个低 就挪动那个 为了两个指针相遇的时候不会有冲突 而且相遇的时候也是低的一边决定水位
然后被挪动的一边判断当前高度和这边max的值 假如更高了 就update max, 否则就填一部分 填的是当前高度这个unit的上面和相应这边max之间的水
public class Solution { public int trap(int[] height) { int left = 0; int right = height.length - 1; int maxLeft = 0; int maxRight = 0; int res = 0; while ( left <= right ){ if ( height [left] <= height [right] ){ if ( height[ left ] > maxLeft ) maxLeft = height [ left ]; else res += maxLeft - height[left]; left ++; } else { if ( height [ right ] > maxRight ) maxRight = height [ right ]; else res += maxRight - height[right]; right --; } } return res; }}
还有一个办法是用stack 并没有研究明白
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