https://choy-mutao.github.io/dsa_1/code_template/Recent content in 代码模板 on MUHugozh-cnhttps://choy-mutao.github.io/dsa_1/code_template/%E5%8F%8C%E6%8C%87%E9%92%88/Mon, 01 Jan 0001 00:00:00 +0000https://choy-mutao.github.io/dsa_1/code_template/%E5%8F%8C%E6%8C%87%E9%92%88/当只有一个输入, 从该输入的两端遍历 int fn(vector<int>& arr) { int left = 0; // left pointer index int right = int(array.size()) - 1; // right pointer index int ans = 0; while(left < right) { if(true /* condition */) { // ... left++; } else { // ... right--; } } return ans; } 有两个输入, 两个都需要遍历 public int fn(int[] arr1, int[] arr2) { int i = 0; // pointer at arr1 int j = 0; // pointer at arr2 int ans = 0; while(i < arr1.