Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container.
面试中比较常问的一道题,开始还以为是选择两个极大值点和一个极小值点,理解错了
第一种解法,直接暴力,时间复杂度O(n^2)
第二种解法,有点类似一次快排的过程
#include <iostream> #include <vector> #include <algorithm> using namespace std; int maxArea(vector<int> &height){ int maxWaterArea = 0; int left = 0, right = height.size()-1; while (left < right) { maxWaterArea = max((right-left)*min(height[left],height[right]),maxWaterArea); if (height[left] < height[right]) left++; else right--; } return maxWaterArea; }
关于如何证明最大值会出现在上述移动的方法中?可以用反证法证明。
当前首尾指针分别是i和j,其中A[i] < A[j],那么移动i。
假设有个k,在i和j之间,A[k]和A[i]组成容器所容纳的水最多,那么势必A[k]>A[j]
然而,A[i]和A[k]组成的容器的容量(k-i)*min(A[i],A[k]),即(k-i)*A[i],小于A[i]与A[j]所组成的容器大小:(j-i)*A[i]。
所有要移动i,而不是移动j