leetcode 40. Combination Sum II

与上一题差不多。但结果不能重复，我们可以做一个去重处理

var combinationSum2 = function (nums, target) {
      if (!Object(nums).length) {
        return [];
      }
      nums.sort()
      var uniq = {}, result = [], n = nums.length, condidate = []
      function backtrack(start, sum) {
        if (sum === 0) {
          var key = condidate + ""
          if (!uniq[key]) {
            result.push(condidate.concat());
            uniq[key] = 1
          }

        } else if (sum > 0) {
          for (var i = start; i < n; i++) {//注意从0开始，不断尝试
            var el = nums[i]
            condidate.push(el) //试探
            backtrack(i+1, sum - el); //递归自身
            condidate.pop(); //不管成功与否，退回上一步
          }
        }
      }
      backtrack(0, target);
      return result;
    };

leetcode 40. Combination Sum II

原文：https://www.cnblogs.com/rubylouvre/p/12045184.html