swap the unordered neighbor, you can prove that after swap the correct order still hold for the previous seriesBinary SearchFind the first value greater then targetSame as Unique pathExcept when init, every thing after obstacle is 0And in the double loop, if obstacle then 0Another outer loop from 3sum, need to check for duplicates in each loophashmap记录inorder值于index的对应然后dfs，preorder第一个为root然后通过map找到root的index，和inStart，inEnd比较的到左右的大小然后再在preorder里找到根据左右大小找到left的root和right的root，如此迭代 public TreeNode build(int pre, int inStart, int inEnd, int[] preorder, HashMap<Integer,Integer> map){ if(pre >= preorder.length || inStart > inEnd) { return null; } TreeNode root = new TreeNode(preorder[pre]); if(inStart == inEnd) { return root; } int index = map.get(preorder[pre]); int leftSize = index - inStart; int rightSize = inEnd - index; int leftRoot = pre + 1; int rightRoot = pre + leftSize + 1; root.left = build(leftRoot, inStart, index-1, preorder, map); root.right = build(rightRoot, index+1, inEnd, preorder, map); return root; }对撞型指针 int left = 0, right = numbers.length - 1; while (left < right) { int sum = numbers[left] + numbers[right]; if (sum < target) { left++; } else if (sum > target) { right--; } else { break; } } return new int[] {left + 1, right + 1};BacktrackingSimiliar to SubsetsUse if (i != idx && nums.get(i) == nums.get(i - 1)) { continue; }to avoid duplicatesbacktracking public void backtracking(List<List<Integer>> res, List<Integer> tmp, int num, int tmpSum, int k, int n) { if (tmp.size() == k && tmpSum == n) { res.add(new ArrayList<Integer>(tmp)); return; } for (int i = num; i <= 9; i++) { if (tmpSum + i <= n) { tmp.add(i); backtracking(res, tmp, i + 1, tmpSum + i, k, n); tmp.remove(tmp.size() - 1); } } }Two pointers因为可能一次要增加两个，所以外层也用while里程while要记录old_j，然后和j比较来确定duplicate的个数 int n = nums.length; int i = 0, j = 0; while (i < n) { int old_j = j; while (j < n && nums[j] == nums[old_j]) { j++; } nums[i++] = nums[old_j]; if (j - old_j > 1) { nums[i++] = nums[old_j]; } if (j == n) { break; } } return i;上下左右4个变量 while(xh > xl && yh > yl){ int x, y; for(x=xl, y=yl; x<xh; x++) res.add(matrix[y][x]); for(x=xh-1, y=yl+1; y<yh; y++) res.add(matrix[y][x]); for(x=xh-2, y=yh-1; x>=xl && y>yl; x--) res.add(matrix[y][x]); for(x=xl, y=yh-2; y>yl && x<xh-1; y--) res.add(matrix[y][x]); xl++; xh--; yl++; yh--; }Two pointers for (i = 0; i < n; i++) { while (j < n && sum < s) { sum += nums[j]; j++; } if (sum >= s) { res = Math.min(res, j - i); } sum -= nums[i]; }counting sort?把值为i的放到i - 1位置上最后遍历，不对应的就是缺失的 for (int i = 0; i < nums.length; i++) { while (nums[i] != i + 1 && nums[i] != nums[nums[i] - 1]) { int tmp = nums[i]; nums[i] = nums[tmp - 1]; nums[tmp - 1] = tmp; } } List<Integer> res = new ArrayList<Integer>(); for (int i = 0; i < nums.length; i++) { if (nums[i] != i + 1) { res.add(i + 1); } }上下左右4个变量 while (left <= right) { // top for (int i = left; i <= right; i++) { matrix[top][i] = cnt; cnt++; } top++; // right for (int i = top; i <= bottom; i++) { matrix[i][right] = cnt; cnt++; } right--; // bottom for (int i = right; i >= left; i--) { matrix[bottom][i] = cnt; cnt++; } bottom--; // left; for (int i = bottom; i >= top; i--) { matrix[i][left] = cnt; cnt++; } left++; }通过循环的询问可以找到candidate，每次询问排除两者之一最后检查一下这个candidate是否valid int candidate = 0; for(int i = 1; i < n; i++){ if(knows(candidate, i)) { candidate = i; } } for(int i = 0; i < n; i++){ if(i != candidate && (knows(candidate, i) || !knows(i, candidate))) { return -1; } } return candidate;Similar to maximum subarray II with left and right array, the only difference is one of them can be 0 (left[0] = 0 and right[n - 1] = 0)Binary Search1. Treat as a 1-D array2. find the last start less then target, then binary search• state: f[x][y]从起点走到x,y的min• function: f[x][y] = min(f[x-1][y], f[x][y-1]) + A[x][y]• intialize: f[i][0] = sum(0 - i) f[0][i] = sum(0 - j)• answer: f[n-1][m-1]k = k%n; reverse(nums,0,n-k-1); reverse(nums,n-k,n-1); reverse(nums,0,n-1);先swap upside down，再symmetric （转置） // first upside down for(int i=0; i<n/2; i++){ for(int j=0; j<n; j++){ int temp = matrix[i][j]; matrix[i][j] = matrix[n-1-i][j]; matrix[n-1-i][j] = temp; } } // then swap symmetrically for(int i=0; i<n; i++){ for(int j=i+1; j<n; j++){ int temp = matrix[i][j]; matrix[i][j] = matrix[j][i]; matrix[j][i] = temp; } }Backtrackingsort the array first, during backtracking, tmpSum, use this if (i != idx && candidates[i] == candidates[i - 1]) { continue; }to avoid duplicates
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