标签：[] col height tco res int points another big
Leetcode: 11
there are two ways to deal with two pointers
one is O(n), two pointers moves from both side
Another is O(2N), two pointer move from the same side
Idea for this, choose the first one and then if there is a smaller one, change that corresponding pointer until bigger that the prior bigger one.
class Solution { public int maxArea(int[] height) { //two pointers -> N //two pointers -> 2*N //6,5,4,7,8,2,4,6,9, //1 2 3 4 5 6 7 8 9 //idea: (i ++) (j--) , for two points pointed by two pointers, there must be one big and one small, change the pointer pointed to small so that mamimize the possibility int i = 0; int j = height.length-1; int max = 0; while(i<j){//if i==j break //compute the area if(height[i]>height[j]){//i bigger int temp = (j-i)*(height[j]); if(temp>max) max = temp; j--; }else { int temp = (j-i)*(height[i]); if(temp>max) max = temp; i++; } } return max; } }
haishi cai
Leetcode 11. Container With Most Water (two pointers)
标签：[] col height tco res int points another big
原文地址：https://www.cnblogs.com/stiles/p/leetcode11.html
