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【力扣】每日三题 -- Day4(数组)

其生 2022-01-20 阅读 40

文章目录

39. 组合总和 – 回溯算法

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        Arrays.sort(candidates);
        backtrack(candidates, target, path, res, 0);
        return res;
    }

    public void backtrack(int[] candidates, int target, List<Integer> path, List<List<Integer>> res, int start) {
        if (target < 0) 
            return;
        if (target == 0) {
            res.add(new ArrayList<>(path));
            return;
        }
        for (int i = start; i < candidates.length; i ++)  {
            if(candidates[i] > target)
                return;
            if(i > 0 && candidates[i] == candidates[i-1])
                continue;

            path.add(candidates[i]);
            backtrack(candidates, target - candidates[i], path, res, i);
            path.remove(path.size() - 1);

        }
    }
}

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64. 最小路径和

class Solution {
    public int minPathSum(int[][] grid) {
        if(grid == null || grid.length == 0 || grid[0].length == 0)
            return 0;
        int m = grid.length;
        int n = grid[0].length;
        int[][] dp = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 && j == 0)
                    dp[i][j] = grid[i][j];
                else if (i == 0)
                    dp[i][j] = grid[i][j] + dp[i][j-1];
                else if (j == 0)
                    dp[i][j] = grid[i][j] + dp[i-1][j];
                else 
                    dp[i][j] = grid[i][j] + Math.min(dp[i-1][j], dp[i][j-1]);
            }
        }
        return dp[m-1][n-1];
    }
}

105. 从前序与中序遍历序列构造二叉树

162. 寻找峰值

在这里插入图片描述

class Solution {
    public int findPeakElement(int[] nums) {
        int left = 0, right = nums.length - 1;
        while(left < right) {
            int m = left + (right - left)/2;
            if( nums[m + 1] < nums[m])
                right = m;//m有可能是峰值
            else
                left = m + 1;
        }
        return left;
    }
}

山脉、峰值问题:
#852 山脉数组的峰顶索引 (数组长度>=3)
#剑指 Offer II 069 山峰数组的顶部 (和852一模一样)
#162 寻找峰值 (数组长度>=1 解法和852就不一样了)
#1095 山脉数组中查找目标值 (考察最为全面,同时考察峰顶索引、二分法-左区间、二分法-右区间三种)
#941 有效的山脉数组 (双指针)
#845 数组中的最长山脉 (双指针)

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